Question

2. As shown in figure, an open beaker, 6 cm in height is filled with liquid benzene at 25°C to within 0.5 cm of the top (evaporated). A gentle breeze of dry air at 25°C and 1 atm is blown by a fan across the mouth of the beaker so that evaporated benzene is carried away by convection after it transfers through a stagnant air layer in the beaker. The vapor pressure of benzene at 25°C is 0.131 atm. The mutual diffusion coefficient for benzene in air at 25°C and 1 atm is 0.0905 cm Is. Compute: (a) The initial rate of evaporation of benzene as a molar flux in mol/cm -s (b) The initial mole-fraction profiles in the stagnant air layer (c) The initial fractions of the mass-transfer fluxes due to molecular diffusion d) The initial diffusion velocities, and the species velocities (relative to stationary coordinates) in the stagnant layer (e) The time in hours for the benzene level in the beaker to drop 2 cm from the initial level, if the specific gravity of liquid Neglect the accumulation of air in the stagnant layer as it Air 1 atm 25 C benzene s 0.874 benzene and Mass transfer0.5 cm reases in height. A-PAUP Interface Liquid Benzene Beaker

Answer 1

Criven deler Ang larm /banfooigem ugning - Benene Diffusinity = 0.0905 auf NATO mass IT PAI = 0.131 auton (repor pressure of bergene) NA - PA PA2 = 0 [- beyene as Gem comed amay by Convention] Let A denotes beyone Beaker B dendes air here only bezene is diffusing into air As is non-diffusing la diffusing through I eveposetion of begene I non-diffusing B] MA Dor P Char-bar) RT Z patm Persoal pressine) = 10tom PAH PBI = 1 0.131 + P si= 1 & bei 1 - 0.121 I= 0.669 atry/ P2 +2 = 1

Prim = PBi-PBI In PBL PBI - 1-0.669 in 1 0.669 Perm= 0.933 / Initial rele of eveposetion of begene . NA = Dan P (Porbr) PT 2 beim = 0.0905X1 (0.131-0) 83.14 X 296X.5*0.933 [ N = 1.03x10-6 ml / cruce a diffusing though Concert Concenprlief A (6) Since et le the case of non-diffusing 3. This is the mole faution profiles in the stagnent al layer Concertations

PA = CART O PT 83.148296 6 = 5.2878106 mol Cma is this due to moleculez diffusion Cakulted by fucks low JA – Dre des JAZ Dm Ci- cm - 0905 4(5.2678156_o) 15. - 2.29 254106 model froses i partion due to of the mass - bansfer nolecuber deffresion 1.038 10 6 2.39252186 = .43

=d) Diffusion inlocuity Vad - J . Vad = 2.39258106 mol/cni-see 5.287 +10b mellem? Vad = 0.1525 con/sec Vm = 1.02x10-6 5.287x106 na o.1940 con/s to the stationary velouty of A subtire Coordineles u= Vad of Un = u525+ 1946 = 0.6473 compren]